Organic Compounds Containing Nitrogen

• $\mathrm{R}-\mathrm{CN} \xrightarrow{\mathrm{Na}(\mathrm{Hg}) / \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}} \mathrm{R}-\mathrm{CH}_2-\mathrm{NH}_2$

• $\mathrm{R}-\mathrm{CN} \xrightarrow[\text { (ii) } \mathrm{H}_2 \mathrm{O}]{\text { (i) } \mathrm{LiAlH}_4} \mathrm{R}-\mathrm{CH}_2-\mathrm{NH}_2$

• $\mathrm{R}-\mathrm{CN} \xrightarrow{\mathrm{Sn}+\mathrm{HCl}} \mathrm{R}-\mathrm{CHO}$

• $\mathrm{R}-\mathrm{CN} \xrightarrow{\mathrm{H}_2 / \mathrm{Ni}} \mathrm{R}-\mathrm{CH}_2-\mathrm{NH}_2$

$ \mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl} \xrightarrow{\mathrm{AgCN}} \underset{\text { (foul smell compound) }}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC}(\mathrm{Z})} $

$ \begin{aligned} & \text { So, } X=\mathrm{AgCN} \\ & Z=\mathrm{C}_2 \mathrm{H}_5 \mathrm{NC} \end{aligned} $

The ortho and para isomers are separated by fractional distillation under reduced pressure.

o-isomer, (M.P. $\rightarrow-4^{\circ} \mathrm{C}$, B.P. $\rightarrow 222^{\circ} \mathrm{C}$ )

p-isomer, (M.P. $\rightarrow 54^{\circ} \mathrm{C}$, B.P. $\rightarrow 238^{\circ} \mathrm{C}$ )

$ \text { Grignard reagents react with amines to extract proton and convert into corresponding hydrocarbon. } $

$ \text { Oxidation of p-nitrotoluene with acidic } \mathrm{KMnO}_4 \text { converts into p-nitrobenzoic acid. } $

$ \text { because of -I effect and -M effect, making p -nitrobenzoic acid as stronger acid than benzoic acid. } $

Reduction of p-nitrotoluene with $\mathrm{Sn} / \mathrm{HCl}$ followed by neutralization reduces the ( $-\mathrm{NO}_2$ ) to an amino group ( $-\mathrm{NH}_2$ ), forming p-toluidine.

$ \text { i.e., } \mathrm{p} \text {-methylaniline is more basic than aniline due to the }+\mathrm{I} \text { effect and hyperconjugation. } $

Lower is the value of $\mathrm{pK}_{\mathrm{b}}$, higher is the basicity

Also aliphatic amines are stronger bases than aromatic amines.

pKb : Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine

Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine

Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278 K .

Nitrous acid is produced in the reaction mixture by reaction of $\mathrm{NaNO}_2$ with HCl.

$\mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2+\mathrm{NaNO}_2+2 \mathrm{HCl} \xrightarrow{273-278 \mathrm{~K}} \mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{N}_2^{+} \stackrel{-}{\mathrm{Cl}}+\mathrm{NaCl}+2 \mathrm{H}_2 \mathrm{O}$

Benzene diazonium chloride decomposes easily in the dry state

lodobenzene is prepared by shaking benzene diazonium salt with KI because direct insertion of iodine into benzene ring is difficult

For positive carbylamine test, there must be the presence of primary amine.

$\bullet$ Aniline does not undergo Friedel-Crafts alkylation reaction due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst.

$\bullet$ Aniline (aromatic primary amine) cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide.

Benzene reacts with chlorine in presence of anhydrous FeCl₃ to give chlorobenzene.

$\bullet$ Primary aliphatic amines react with HNO₂ and give unstable diazonium salt which turns into alcohol

$R - N{H_2} + HN{O_2}\buildrel {} \over \longrightarrow \left[ {R - N_2^ + - C{l^ - }} \right]\buildrel {{H_2}O} \over \longrightarrow ROH + {N_2} + HCl$

$\bullet$ Primary aromatic amines reacts with HNO₂ and give stable diazonium salt which are stable at 273 to 278 K.

${C_6}{H_5} - N{H_2} + HN{O_2}\buildrel {273 - 278\,K} \over \longrightarrow {C_6}{H_5}N_2^ + C{l^ - }$

We do not get tert-butyl amine using Gabriel phthalimide synthesis. Gabriel phthalimide synthesis only gives primary aliphatic-amines formation of tert-butyl amine follow S$_N$1, mechanism due to formation of stable cation. Thus, assertion is incorrect but reason is correct. Hence, option (d) is the correct.

Both assertion and reason are correct and reason is the correct explanation of the assertion. In nitration of benzene with nitric acid, sulphuric acid acts as catalyst. It helps in the formation of electrophile, i.e. nitronium ion NO$_2^+$.

The reaction is shown below :

$\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \mathrm{NO}_2^{+}+2 \mathrm{HSO}_4^{-}+\mathrm{H}_3 \mathrm{O}^{+}$